What is the empirical formula for a compound that is 50.8 zinc 16.0 phosphorus and 33.2 oxygen

1 answer:

5 0

1)number mole in
50.8 g Zn *(1 mol Zn/65.4g)=0.777 mol Zn
2)number mole in
16.0 g P*(1 mol P/31.0 g)= 0.516 mol P
3)number mole in
33.2 g O*(1 mol O/16.0 g) = 2.075 mol O

0.777 mol Zn : 0.516 mol P : 2.075 mol O =
= (0.777 mol Zn/0.516) : (0.516 mol P/0.516) : (2.075 mol O/0.516)=
=1.5 mol Zn : 1 mol P : 4 mol O=3 Zn : 2 P : 8O
Zn3P2O8
or
the same is Zn3(PO4)2

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